1.4 Conditional probability
1 (a)
$$ \begin{array}{r c l} P(A|B) & = & \dfrac{P(A \cap B)}{P(B)} \\[0.5em] & = & \dfrac{0.01 + 0.12 + 0.05}{ 0.01 + 0.02 + 0.05 + 0.11 + 0.06 + 0.08 + 0.13 } \\[0.5em] & = & \dfrac{0.18}{0.46} \\[0.5em] & = & 0.3913 \end{array} $$(b)
$$ \begin{array}{r c l} P(C|A) & = & \dfrac{P(C \cap A)}{P(C)} \\[0.5em] & = & \dfrac{ 0.08 + 0.02 + 0.04 + 0.05 }{ 0.08 + 0.02 + 0.04 + 0.05 + 0.07 + 0.11 + 0.11 } \\[0.5em] & = & \dfrac{0.19}{0.48} \\[0.5em] & = & 0.3958 \end{array} $$(c)
$$ \begin{array}{r c l} P(B|A \cap B) & = & \dfrac{P(B \cap (A \cap B))}{P(A \cap B)} \\[0.5em] & = & \dfrac{P(B \cap A \cap B)}{P(A \cap B)} \\[0.5em] & = & \dfrac{P(B \cap B \cap A)}{P(A \cap B)} \\[0.5em] & = & \dfrac{P(B \cap A)}{P(A \cap B)} \\[0.5em] & = & \dfrac{P(A \cap B)}{P(A \cap B)} \\[0.5em] & = & 1 \end{array} $$(d) Because
$$ \begin{array}{r c l} P(B|A \cup B) & = & \dfrac{P(B \cap (A \cup B))}{P(A \cup B)} \\[0.5em] & = & \dfrac{P((B \cap A) \cup (B \cap B))}{P(A \cup B)} \\[0.5em] & = & \dfrac{P((B \cap A) \cup (B))}{P(A \cup B)} \\[0.5em] & = & \dfrac{P(B \cap A) - P(B) - P((B \cap A) \cap B)}{P(A \cup B)} \\[0.5em] & = & \dfrac{P(B \cap A) - P(B) - P(B \cap A \cap B)}{P(A \cup B)} \\[0.5em] & = & \dfrac{P(B \cap A) - P(B) - P(A \cap B \cap B)}{P(A \cup B)} \\[0.5em] & = & \dfrac{P(B \cap A) - P(B) - P(A \cap B)}{P(A \cup B)} \\[0.5em] & = & \dfrac{P(B \cap A) - P(B) - P(A \cap B)}{P(A \cup B)} \\[0.5em] \end{array} $$and $P(B \cap A)$ $=$ $0.01$ $+$ $0.02$ $+$ $0.05$ $=$ $0.08$, $P(A)$ $=$ $0.07$ $+$ $0.05$ $+$ $0.01$ $+$ $0.08$ $+$ $0.02$ $+$ $0.04$ $+$ $0.05$ $=$ $0.32$ and $P(B)$ $=$ $0.46$, then
$$ \begin{array}{r c l} \dfrac{P(B \cap A) + P(B) - P(A \cap B)}{P(A \cup B)} & = & \dfrac{0.08 + 0.46 - 0.08}{ \begin{array}{l} \phantom{+} 0.07 + 0.05 + 0.01 \\ + 0.08 + 0.02 + 0.04 \\ + 0.05 + 0.11 + 0.06 \\ + 0.08 + 0.13 \end{array} } \\[0.5em] & = & \dfrac{0.46}{0.7} \\[0.5em] & = & 0.6571 \end{array} $$(e)
…
(f) Because
$$ \begin{array}{r c l} P(A \cap B|A \cup B) & = & \dfrac{P((A \cap B) \cap (A \cup B))}{P(A \cup B)} \\[0.5em] & = & \dfrac{P(A \cup B)}{P(A \cup B)} \\[0.5em] & = & 1, \end{array} $$because $(A \cap B)$ $\in$ $(A \cup B)$.